Integral of e^(x^2 + y^2) dx dy over x from 0 to sqrt(25 - y^2) and over y from -5 to 5. (a)...


{eq}\displaystyle \int_{-5}^{5} \int_{0}^{\sqrt{25 - y^2}} e^{(x^2 + y^2)}dx \, dy {/eq}

(a) Sketch the region of integration.

(b) Change to polar coordinates.

(c) Evaluate the integral.

Polar Coordinates:

When faced with a double integral, recall that not only can we switch the order of integration but we can switch coordinate systems entirely. For the situation above, it will be more convenient to use polar coordinates.


{eq}x = r \cos \theta {/eq}

{eq}y = r \sin \theta {/eq}

{eq}r^2 = x^2+y^2 {/eq}

{eq}\theta = \tan^{-1} \frac{y}{x} {/eq}

{eq}dA = r\ dr\ d\theta {/eq}

Answer and Explanation:

Part A

From the limits on the integral, we see that the region of integration is {eq}0 \leq x \leq \sqrt{25-y^2} {/eq} when {eq}y \in [-5,5] {/eq}....

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Learn more about this topic:

Graphing Functions in Polar Coordinates: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 1

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